# Find the integrals of the functions$\sin^2(2x+5)$

 Solution: $\frac{1}{2}x - \frac{1}{8}sin(4x+10)+k$

Toolbox
• Anytime you see a problem that is has a $\sin^2x$ function, you cannot integrate it directly. Instead, you can use the following trignometric identity: $sin^2x = \frac{1}{2}(1 – cos2x)$ This converts the $\sin^2x$ function into something you can work with.
• $\int cos(\alpha\;x+\beta) \; dx = \frac{1}{\alpha} sin(\alpha \; x + \beta)$

Step-1

• Let us use the information from the Toolbox above: $\int \; sin^2(2x + 5) \; dx = \frac{1}{2}(1 – cos2\;(2x+5) \; dx$.
• This identity expands to $\frac{1}{2}(1 – cos(4x+10) \; dx$

Step-2

• $\int \frac{1}{2} (1-cos(4x+10)) dx = \int \frac{1}{2}dx - \frac{1}{2} \int cos(4x+10)dx$
• Now, $\int\frac{1}{2}\;dx = \frac{1}{2}$. We need to evaluate $\int cos(4x+10)\; dx$
Step-3
• Hint: Remember: $\int cos(\alpha\;x+\beta) \; dx = \frac{1}{\alpha}sin(\alpha\;x+b)$
• Therefore. $\int cos(4x+10)\; dx = \frac{1}{4}sin(4x+10)$

Step-4

• Evaluating the entire term above, we get $\int \; sin^2(2x + 5) \; dx = \frac{1}{2}x - \frac{1}{2}\;\frac{1}{4}sin(4x+10) + k (Constant)$
• Rewriting, we get the solution as $\frac{1}{2}x - \frac{1}{8}sin(4x+10) + k$

edited Dec 6, 2012